Building an algebraic closure
October 24, 2011 1 Comment
I need to show you how you can add the root of an element in an overly complicated and abstract way. If you’re thinking of , or something like that, and you want to add a root of , you can just tack on to make . But what if you don’t know a priori what the root is that you want to add? Or what if the field isn’t something you understand well? What if it’s one of those confusing finite fields?
Disclaimer: I’m going to use a fair amount of ring theory without even really defining the terms. If you feel lost, don’t worry about it. I’ve included links so you can try to learn what you want. This sort of thing will be restricted to this post, so if you don’t understand, you’ll still be fine for later posts.
Let’s try to be very general. is a field, and is an irreducible polynomial (meaning it can’t be factored) with coefficients in . That is, . What I can do is let be a variable, and consider the ring . This is exactly the same as (just replacing with ). The reason I switch is because I don’t want to confuse the two. I want to always represent a variable. In a moment, will be the root of our polynomial.
Now I’m going to take the ideal generated by in . We write this as for reasons I don’t totally understand. In any event, this is a prime ideal. That means that if is in the ideal, then one of and are as well. The reason I know this is because we chose to be irreducible. The analogous situation in the integers is that if is some multiple of a prime , then at least one of and are divisible by this prime.
But what’s even sweeter is that in this ring, is a root of the polynomial . Since modding out by essentially says “I declare that is zero!”
What’s even even sweeter is that is a subfield of this new field .
This whole thing was a bit nasty, but it means that we can take any field, and add the root of a polynomial, even if we don’t know a priori what that root is.